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Find the values of $p$ for which the series $\sum_{n=2}^\infty \frac{1}{n(\ln n)^p}$ is convergent.

I know that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent if $p>1$ and divergent if $p\leq1$.

So with the series in question, I can't just say "when $p>1$" right? Because the denominator is $n(\ln n)^p$ not just $n^p$ as the formula states. So how should I go about tackling this problem? I know that doing a u-sub for $\ln n$ would work because $du$ would equals $\frac{dn}{n}$ but what would I do with the $p$?

Thanks for the help.

user1729
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Logan
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    You can use the integral test – Brandon Jan 17 '14 at 05:24
  • @Brandon Okay, but how would I determine what p has to be from the integral test other than just plugging in numbers and guessing? – Logan Jan 17 '14 at 05:27
  • @Brandon Okay, so basically the ln(x) disappears/doesn't really matter? Is this because of a comparison test or something else? I do know that 1/x^p converges if p>1. – Logan Jan 17 '14 at 05:36
  • I have voted to reopen this question as the "target" question is specifically about the $p>1$ case. The question here is more general. – user1729 Aug 25 '20 at 13:21

2 Answers2

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$\sum_{n=2}^\infty \frac{1}{n(\ln n)^p}$ You can use the integral test

$\int_{2}^\infty \frac{1}{x(\ln x)^p}dx=\lim_{A \to \infty}\int_{2}^A\frac{1}{x(\ln x)^p}dx=\lim_{A \to \infty}\int_2^A (\ln x)^{-p}d (\ln x)$

Let $p>1$ then $\lim_{A \to \infty}\int_2^A (\ln x)^{-p}d (\ln x)$ $=\lim_{A\to \infty}\left[\frac{(\ln A)^{1-p}}{1-p}-\frac{(\ln 2)^{1-p}}{1-p}\right]$ $=-\frac{(\ln 2)^{1-p}}{1-p}$ because $1-p<0$ and $\ln A \to \infty$ as $ A \to \infty$.

Therefore in this case integral and seris are convergent.

Let $p<1$ then $\lim_{A \to \infty}\int_2^A (\ln x)^{-p}d (\ln x)$ $=\lim_{A\to \infty}\left[\frac{(\ln A)^{1-p}}{1-p}-\frac{(\ln 2)^{1-p}}{1-p}\right]$ $=\infty$ because $1-p>0$ and $\ln A \to \infty$ as $ A \to \infty$.

Therefore in this case integral and seris are divergent.

Let $p=1$ then $\lim_{A \to \infty}\int_2^A (\ln x)^{-1}d (\ln x)$ $=\lim_{A\to \infty}\left[\ln(\ln A)-\ln(\ln 2)\right]$ $=\infty$ because $\ln A \to \infty$ as $ A \to \infty$.

Therefore in this case integral and seris are divergent.

kmitov
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  • Awesome! Thanks for such a thorough answer! Can you please explain to me why/how the integral changed for when p=1? – Logan Jan 17 '14 at 05:49
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    It is written above. If $p=1$ the integral is $\int \frac{1}{x \ln x}dx=\int \frac{1}{\ln x} d \ln x=\ln (\ln x).$ – kmitov Jan 17 '14 at 08:02
  • Oh okay I get it now. Thanks so much! – Logan Jan 17 '14 at 16:02
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An alternative solution would be to use Cauchy's condensation test. Namely, the function $f(n)= n^{-p} \ln(n)^{-p}$ is eventually non-increasing for all $p \in \mathbb{R}$. Thus, it suffices to consider the series

$$\sum_{n=1}^{\infty} 2^nf(2^n) = \frac{1}{\ln(2)^p}\sum_{n=1}^{\infty} \frac{1}{2^{n(p-1)} n^p}.$$

And we see that we have convergence if and only if $p>1$.

wolverine03
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